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Find all real numbers $k$ for which the equation $(k-5)x^2-kx+5=0$ has exactly one real solution.

Sagot :

lhannearaujo

Answer:

The equation has exactly one real solution when k=10.

Step-by-step explanation:

A quadratic equation is represented by the form:

[tex]ax^{2} + bx+ c=0 , where:\\\\ a\neq 0\\ and \\ a,b, c= coeficients[/tex]

It is possible to determine the number of solutions of quadratic equation from the discriminant (D). The formula for the discriminant is:

[tex]D=b^{2} -4ac[/tex]

When:

  • [tex]D=b^{2} -4ac > 0[/tex] -  the quadratic equation will have 2 solutions.
  • [tex]D=b^{2} -4ac < 0[/tex] -  there are no solutions for the quadratic equation.
  • [tex]D=b^{2} -4ac = 0[/tex] -  the quadratic equation will have 1 solution.

For solving this question you should follow the next steps.

  • Step 1 -  Apply the formula for the discriminant

[tex]D=b^{2} -4ac[/tex], for the given equation, we have:

a= k-5

b=-k

c=5

Therefore,

[tex]D=b^{2} -4ac\\\\ D= (-k)^2-4*(k-5)*5\\\\ D=k^{2} -20k+100[/tex]

  • Step 2 -  Apply the condition for quadratic equation to have one solution

 The quadratic equation will have one solution when D=0. Thus,

[tex]D=k^{2} -20k+100=0[/tex]

  • Step 3 -  Find the values for k

For this, you should solve the equation:  [tex]k^{2} -20k+100=0[/tex]

We can rewrite [tex]k^{2} -20k+100=0[/tex] as  [tex](k-10)^2=0[/tex]

Therefore:

[tex](k-10)^{2} =0\\ \sqrt{(k-10)^{2}} =\sqrt{0} \\ k-10=0\\ k=10[/tex]

This equation has exactly one real solution when k=10.

Learn more about number of solution for a quadratic equation here:

https://brainly.com/question/7784687

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